Exploring Periodic Data

Exploring Periodic DataExploring Periodic DataALGEBRA 2 LESSON 13-1ALGEBRA 2 LESSON 13-1

(For help, go to Lesson 2-1.)

Determine whether each relation is a function.

1. {(2, 4), (1, 3), (–3, –1), (4, 6)} 2. {(2, 6), (–3, 1), (–2, 2)}

3. {(x, y)| x = 3} 4. {(x, y)| y = 8}

5. {(x, y)| x = y2} 6. {(x, y)| x2 + y2 = 36}

7. {(a, b)| a = b3} 8. {(w, z)| w = z – 36}

1. {(2, 4), (1, 3), (–3, –1), (4, 6)}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

2. {(2, 6), (–3, 1), (–2, 2)}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

3. {(x, y)| x = 3}; no, this is not a function because it is a vertical line and fails the vertical line test.

4. {(x, y)| y = 8}; yes, this is a function because it is a horizontal line and passes the vertical line test.

Solutions

5. {(x, y)| x = y2}; no, this is not a function because an element of the domain is paired with more than one element in the range.Example: 4 = 22 and 4 = (–2)2

6. {(x, y)| x2 + y2 = 36}; no, this is not a function because it is a circle and fails the vertical line test.

7. {(a, b)| a = b3}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

8. {(w, z)| w = z – 36}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

Solutions (continued)

Analyze this periodic function. Identify one cycle in two

different ways. Then determine the period of the function.

Each cycle is 7 units long. The period of the function is 7.

Begin at any point on the graph.Trace one complete pattern.

The beginning and ending x-values of each cycle determine the period of the function.

Determine whether each function is or is not periodic. If it is,

find the period.

The pattern of y-values in one section repeats exactly in other sections. The function is periodic.

Find points at the beginning and end of one cycle.

Subtract the x-values of the points: 2 – 0 = 2.

The pattern of the graph repeats every 2 units, so the period is 2.

(continued)

The pattern of y-values in one section repeats exactly in other sections. The function is periodic.

Find points at the beginning and end of one cycle.

Subtract the x-values of the points: 3 – 0 = 3.

The pattern of the graph repeats every 3 units, so the period is 3.

Find the amplitudes of the two functions in

Additional Example 2.

The amplitude of the function is 2.

a. amplitude = (maximum value – minimum value) Use definition of amplitude.

= [2 – (–2)] Substitute.12

= (4) = 2 Subtract within parentheses and simplify.12

(continued)

The amplitude of the function is 3.

b. amplitude = (maximum value – minimum value) Use definition of amplitude.

= [6 – 0] Substitute.12

= (6) = 3 Subtract within parentheses and simplify.12

The oscilloscope screen below shows the graph of the

alternating current electricity supplied to homes in the United States.

Find the period and amplitude.

1 unit on the t-axis = s1

(continued)

The maximum value of the function is 120, and the minimum is –120.

One cycle of the electric current occurs from 0 s to s.1

period = – 0 Use the definitions.

= Simplify.

amplitude = [120 – (–120)]

= (240) = 120

The amplitude is 120 volts.The period of the electric current is s.1

pages 699–702 Exercises

1. x = –2 to x = 3, x = 2 to x = 7; 5

2. x = 0 to x = 4, x = 5 to x = 9; 4

3. x = 0 to x = 4, x = 2 to x = 6; 4

4. not periodic

5. periodic; 12

6. not periodic

7. not periodic

8. periodic; 8

9. periodic; 7

16. a. y

17. Answers may vary. Sample: Yes; average monthly temperatures for three years should be cyclical due to the variation of the seasons.

18. Answers may vary. Sample: No; population usually increases or decreases but is not cyclical.

19. Answers may vary. Sample: Yes; traffic that passes through an intersection should be at the same levels for the same times of day for two consecutive work days.

20. 60 beats per min

21. a. 1 sb. 1.5 mV

22. Check students’ work.

23. 3, –3, 4;

24. 5, 0, 8;

25. 4, –4, 8;

26. 1 yr

27. 2 weeks

28. 3 months

29. 1 hour

30. 1 day

31. 2, 2, 2

32. a. 67

33. a.

b. 5 s, 1 ftc. Answers may vary. Sample: about 1 s

34. a. 24.22 daysb. 0.78 dayc. 0.22 dayd. Answers may vary. Sample: The calendar year

is meant to predict events in the solar year. Keeping the difference between the two minimal is necessary for the calendar year to be useful.

40. [4] 64 seconds. The first two functions are at the beginning of their cycles together every 6 • 7 = 42 seconds: 42, 84, 126, . … The third function is at the beginning of its cycle every 8 seconds, starting at 20 seconds: 20, 28, 36, 44, 52, 60, 68, 76, 84, . … The three functions are all at the beginning of their cycles at 84 seconds, which is 64 seconds after the third function begins.

[3] minor error in calculation

[2] incomplete explanation

[1] answer only, with no explanation

39. [2] A period is the length of 1 cycle,

Then xm = n, or x = .

The period is seconds.

[1] answer only with no explanation

n secondsm cycles

x seconds1 cycle

42. 11, 13; an = 2n – 1; explicit or a1 = 1, an = an – 1 + 2; recursive

43. 14, 16; an = 2n + 2; explicit or a1 = 4, an = an – 1 + 2; recursive

44. 38, 51; a1 = 3, an = an – 1 + 2n – 1;

recursive or an = n2 + 2; explicit

45. (x + 3)2 = 20(y – 2)

46. – = 1

47. – = 1

(y + 3)2

4(x – 5)2

(y + 2)2

16(y – 1)2

1. Determine whether the function shown is periodic.

2. Determine whether the function shown is periodic.

Find the period and the amplitude ofthe periodic functions shown.

6; 1.5

3; 0.5

Angles and the Unit CircleAngles and the Unit CircleALGEBRA 2 LESSON 13-2ALGEBRA 2 LESSON 13-2

(For help, go to page 54.)

For each measure, draw an angle with its vertex at the origin of the coordinate plane. Use the positive x-axis as one ray of the angle.

1. 90° 2. 45° 3. 30°

4. 150° 5. 135° 6. 120°

Solutions

1. 2. 3.

4. 5. 6.

Find the measure of the angle.

Since 90 + 60 = 150, the measure of the angle is 150°.

The angle measures 60° more than a right angle of 90°.

Sketch each angle in standard position.

a. 48° b. 310° c. –170°

The Aztec calendar stone has 20 divisions for the 20 days in

each month of the Aztec year. An angle on the Aztec calendar shows

the passage of 16 days. Find the measures of the two coterminal

angles that coincide with the angle.

To find a coterminal angle, subtract one full rotation.

288° – 360° = –72°

Two coterminal angle measures for an angle on the Aztec calendar that show the passage of 16 days are 288° and –72°.

The terminal side of the angle is of a full rotation from the initial side.1620

• 360° = 288°1620

Find the cosine and sine of 135°.

Use a 45°-45°-90° triangle to find sin 135°.

From the figure, the x-coordinate of point A

is – , so cos 135° = – , or about –0.71. 22

opposite leg = adjacent leg

0.71 Simplify.

= Substitute. 22

The coordinates of the point at which the terminal side of a 135° angle intersects are about (–0.71, 0.71), so cos 135° –0.71 and sin 135° 0.71.

Find the exact values of cos (–150°) and sin (–150°).

Step 1: Sketch an angle of –150° in standard position. Sketch a unit circle.

x-coordinate = cos (–150°)y-coordinate = sin (–150°)

Step 2: Sketch a right triangle. Place the hypotenuse on the terminal side of the angle. Place one leg on the x-axis. (The other leg will be parallel to the y-axis.)

(continued)

The triangle contains angles of 30°, 60°, and 90°.

Step 3: Find the length of each side of the triangle.

hypotenuse = 1 The hypotenuse is a radius of the unit circle.

shorter leg = The shorter leg is half the hypotenuse.12

32longer leg = 3 = The longer leg is 3 times the shorter leg.

Since the point lies in Quadrant III, both coordinates are negative. The longer leg lies along the x-axis, so

cos (–150°) = – , and sin (–150°) = – .

12. 25°

13. 215°

14. 315°

15. 4°

16. 140°

17. 150°

18. 55°

19. 180°

20. 220°, –140°

1. –315°

2. –135°

3. 240°

4. 115°

5. –110°

6. –340°

31. 0.71, –0.71

32. –0.87, 0.50

33. –0.09, –1.00

34. 0.98, –0.17

35. –0.90, 0.44

36. 0.00, 1.00

37–44. Answers may vary. Samples:

37. 405°, –315°

38. 235°, –485°

39. 45°, –315°

40. 40°, –320°

41. 275°, –445°

42. 295°, –65°

43. 573°, –147°

44. 303°, –417°

45. II

46. III

47. negative x-axis

48. IV

49. positive x-axis

21. , – ; 0.50, –0.87

22. – , – ; –0.71, –0.71

23. , – ; 0.87, –0.50

24. – , ; –0.50, 0.87

25. , ; 0.87, 0.50

26. , – ; 0.71, –0.71

27. , – ; 0.87, –0.50

28. – , ; –0.71, 0.71

29. 1.00, 0.00

30. 0.85, 0.53

50. a.

b. IIc. If the terminal side of an

angle is in Quadrants I or II, then the sine of the angle is positive; otherwise it is not. If the terminal side of an angle is in Quadrants I or IV, then the cosine of the angle is positive; otherwiseit is not.

51. a. 0.77, 0.77, 0.77b. The cosines of the three angles

are equal because the angles are coterminal.

52. The x-coordinate of the point on the ray defined by angle is equal to cos ; similarly for the y-coordinate and sin . The angles 0°, 180°, and 360° lie on the x-axis, and thus their sines are all 0 and their cosines are ±1. The angles 90° and 270° lie on the y-axis, so their cosines are 0 and their sines are ±1.

53. , 32

57. , –

58. , –

59. – ,

60. Answers may vary. Sample: 30°, 150°, –210°, 390°

54. – ,

55. – , –

56. , –

61. No; yes; if the sin and cos are both negative, is in Quadrant III. –120° is in Quadrant III.

62. a. Check students’ work.b. –20°

67. [2]

The terminal side forms an angle of 30° with the negative x-axis. Using

the unit circle, x = – and y = .

So cos (–210°) = – .

[1] answer only, with no work shown

68. [4]

The terminal side forms an angle of 45° with the negative x-axis,

so sin(–135°) = – and cos(–135°) = – .

Then [sin (–135°)]2 + [cos (–135°)]2 =

– + – = + = = 1.

(OR a convincing argument using x2 + y2 = 1)[3] one computational error[2] incomplete explanation with correct answer[1] answer only, with no work shown

69. periodic; 3

70. not periodic

71. periodic; 6

74. (0, 2 5 ), (0, –2 5)

75. (0, 5 5 ), (0, –5 5)

76. ( 85, 0), (– 85, 0)

77. ( 145, 0), (– 145, 0)

Sketch each angle in standard position. Use a right triangle to find the exact values of the cosine and sine of the angle.

1. 45° 2. –120°

3. What angle is less than 360° and coterminal with 45°?

12;– 3

–315°

Radian MeasureRadian Measure

(For help, go to page 870.)

ALGEBRA 2 LESSON 13-3ALGEBRA 2 LESSON 13-3

Find the circumference of a circle with the given radius or diameter. Round your answer to the nearest tenth.

1. radius 4 in. 2. diameter 70 m

3. radius 8 mi 4. diameter 3.4 ft

5. radius 5 mm 6. diameter 6.3 cm

Radian MeasureRadian MeasureALGEBRA 2 LESSON 13-3ALGEBRA 2 LESSON 13-3

1. C = 2 r = 2 (4 in.) 25.1 in.

2. C = d = (70 m) 219.9 m

3. C = 2 r = 2 (8 mi) 50.3 mi

4. C = d = (3.4 ft) 10.7 ft

5. C = 2 r = 2 (5 mm) 31.4 mm

6. C = d = (6.3 cm) 19.8 cm

Solutions

a. Find the radian measure of angle of 45°.

Write a proportion.45°

180° =r radians radians

An angle of 45° measures about 0.785 radians.

Write the cross-products.45 • = 180 • r

Divide each side by 45.r = 45 •180

= 0.785 Simplify.4

(continued)

= 390° Simplify.

b. Find the degree measure of .6

Write a proportion.613

radians= d°

• 180 = • d Write the cross-product.6

d = Divide each side by .13 • 180

6 •1

An angle of radians measures 390°.6

a. Find the degree measure of an angle of – radians.23

= –270°

An angle of – radians measures –270°.2

– radians • = – radians •2

3 180°radians 2

3 180°radians1

90Multiply by

180°radians .

(continued)

b. Find the radian measure of an angle of 54°.

54° • radians = 54° • radians Multiply by radians.180° 180° 180°3

103 radians= Simplify.

An angle of 54° measures radians.103

Draw the angle.

Find the exact values of cos and sin .radians3 radians3

radians • = 60° Convert to degrees.3180°radians

Complete a 30°-60°-90° triangle.

The hypotenuse has length 1.

radians3Thus, cos =12

and sin radians3 = . 32

The shorter leg is the length of the hypotenuse, and the longer leg is 3 times the length of the shorter leg.

Use this circle to find length s to the nearest tenth.

s = r Use the formula.

The arc has length 22.0 in.

= 7 Simplify.

22.0 Use a calculator.

= 6 • Substitute 6 for r and for .7 6

Another satellite completes one orbit around Earth every 4 h.

The satellite orbits 2900 km above Earth’s surface. How far does the

satellite travel in 1 h?

Since one complete rotation (orbit) takes 4 h, the satellite completes of a rotation in 1 h.

Step 1: Find the radius of the satellite’s orbit.

r = 6400 + 2900 Add the radius of Earth and the distance

from Earth’s surface to the satellite.

= 9300

(continued)

The satellite travels about 14,608 km in 1 h.

Step 2: Find the measure of the central angle the satellite travels through in 1 h.

= • 2 Multiply the fraction of the rotation by the number of radians in one complete rotation.

= • Simplify.

Step 3: Find s for = .

s = r Use the formula.

= 9300 • Substitute 9300 for r and for .

14608 Simplify.

1. – , –5.24

2. , 2.62

3. – , –1.57

4. – , –1.05

5. , 2.79

6. , 0.35

7. 540°

8. 198°

9. –120°

10. –172°

11. 90°

12. 270°

16. 0, 1

17. – ,

18. – ,

19. 0, –1

20. 3.1 cm

21. 10.5 m

22. 51.8 ft

23. 25.1 in.

24. 4.7 m

25. 43.2 cm

26. 107 in.

27. 32 ft

28. a. 11,048 kmb. 33,144 kmc. 27,620 kmd. 276,198 kme. 18.1 h

29. 42.2 in.

30. a. 15°, radians

b. 1036.7 mic. 413.6 mi

31. III

32. II

33. positive y-axis

34. II

35. negative x-axis

36. III

0.71, –0.71

–0.50, –0.87

0.00, 1.00

1.00, 0.00

–0.87, –0.50

0.81, –0.5943. a–b.

c. All five triangles are congruent by SSS. All have a hypotenuse of 1 unit, a long leg of about 0.81 unit, and a short leg of 0.59 unit.

cos = 0.81, sin = 0.59;

sin = 0.81, cos = 0.59;

cos = –0.81, sin = 0.59;

cos = –0.81, sin = –0.59;

cos = 0.81, sin = –0.59

45. 11 radians

46. The student forgot to include parentheses.

47. 798 ft; 55°, 665°

48. 23.6 in.; – ,

49. If two angles measured in radians are coterminal, the difference of their measures will be evenly divisible by 2 .

5 53 10

50. 6.3 cm

51. 4008.7 mi

52. – radians

53. – radians

54. radians

55. radians

• 2 r = • 2 r

57. a. 0.5017962; 0.4999646; the first four terms

b. 1 – + – + – . . .

c. 0.951; 18°

62. [2] For a central angle of 1 radian, the length of the intercepted arc is the length of the radius.

68. 9.1, 5.41

69. 12.9, 3.53

70. 30, 8.09

71. x2 + y2 = 64

72. x2 + (y + 5)2 = 16

73. (x – 3)2 + (y – 7)2 = 42.25

74. (x + 8)2 + (y – 4)2 = 9

1. Rewrite each angle measure using the other unit, either degrees or radians.

a. 225°

b. radians

2. A wrench turns through an angle of 1.5 radians. If the wrench is 14 in. long, what is the distance that the end of the wrench moves?

3. A jogger runs 100 m around a circular track with a radius of 40 m. Through what angle does the jogger move? Express your answer in both radians and degrees.

5 4 radians

21 in.

2.5 radians; about 143.2°

The Sine FunctionThe Sine Function

Use the graph. Find the value(s) of each of the following.

1. the period

2. the domain

3. the amplitude

4. the range

The Sine FunctionThe Sine FunctionALGEBRA 2 LESSON 13-4ALGEBRA 2 LESSON 13-4

Solutions

1. the period: 2 units

2. the domain: all real numbers

3. the amplitude: = 1 unit

4. the range: –1 y 1, where y is a real number

<– <–

Use the graph of the sine function.

a. What is the value of y = sin for = 180°?

b. For what other value(s) of from 0° to 360° does the graph of sin have the same value as for = 180°?

The value of the function at = 180° is 0.

When y = 0, = 0° and 360°.

Estimate each value from the graph. Check your estimate

with a calculator.

a. sin 3

sin 3 = 0.1411200081 Use a calculator to check your estimate.

The sine function reaches its median value of 0 at 3.14. The

value of the function at 3 is slightly more than 0, or about 0.1.

(continued)

b. sin 2

sin = 1 Use a calculator to check your estimate.2

The sine function reaches its maximum value of 1 at , so sin = 1.2 2

Use the graph of y = sin 6 .

a. How many cycles occur in this graph? How is the number of cycles related to the coefficient of in the equation?

b. Find the period of y = sin 6 .

The graph shows 6 cycles.

2 ÷ 6 = Divide the domain of the graph by the number of cycles.3

The number of cycles is equal to the coefficient of .

The period of y = sin 6 is .3

This graph shows the graph of y = a • sin for values of

a = and a = 3.34

a. Find the amplitude of each sine curve. How does the value of a affect the amplitude?

In each case, the amplitude of the curve is | a |.

b. How would a negative value of a affect each graph?

When a is negative, the graph is a reflection in the x-axis.

The amplitude of y = 3 • sin is 3.

The amplitude of y = sin is 1, and the amplitude

of y = • sin is .34

Step 1: Choose scales for the y-axis and the x-axis that are about equal ( = 1 unit). On the x-axis, mark one period (4 units).

Step 2: Mark equal spaces through one cycle by dividing the period into fourths.

Step 3: Since the amplitude is 3, the maximum 3 and the minimum is –3. Plot the five points and sketch the curve.

a. Sketch one cycle of a sine curve with amplitude 3

and period 4.

(continued)

b. Use the form y = a sin b . Write an equation with a > 0 for the sine curve in part a.

The amplitude is 3, and a > 0, so a = 3.

The period is 4, and 4 = , so b = .2 b 2

An equation for the function is y = 3 sin x.2

Divide the period into fourths. Using the values of the amplitude and period, plot the zero-max-zero-min-zero pattern.

Sketch one cycle of y = sin 3 .53

| a | = , so the amplitude is .53

b = 3, so there are 3 cycles from 0 to 2 .

= , so the period is .2 b

Sketch the curve.

Find the period of the following sine curve. Then write an

equation for the curve.

According to the graph, one cycle takes 3 units to complete, so the period is 3.

To write the equation, first find b.

period = Use the relationship between

the period and b.

3 = Substitute.2 b

b = Multiply each side by .2 3

2.094 Simplify.

Use the form y = a sin b . An equation for the graph is y = 5 sin .2

2. 0.7

3. 0.9

5. –0.9

6. –0.9

8. 0.1

9. –0.8

10. –1

11. –1

12. –0.7

13. 3; 2,

14. ; 1, 4

15. 2; 3,

y = 2 sin 3

y = sin 2

y = 4 sin

y = 3 sin

y = sin

y = 1.5 sin

28. 2 ; y = 2 sin

29. 2 ; y = –3 sin

30. ; y = sin 2

31. ; y = sin 6

32. ; y = –sin 2

33. 4; y = 3 sin

34. 1; 1, 2

35. 5; 1,

36. ; 1, 2

37. 1; 3, 2

38. 1; 5, 2

39. 2 ; 5, 1

40. a.

b. As a increases, the amplitude of the graph increases.

41. a.

They are reflections of each other in the x-axis.

c. When either a or b is replaced by its opposite, the graph is a reflection of the original graphin the x-axis.

42. a.b. 4

43. a.

b. 0.001c. 880

44. • |a| is the amplitude of the function.

• b is the number of cycles in the

interval 0° to 360°.

• is the period of the function.

The properties relating to number

of cycles and period are affected.

45. , 3.5

360°b

47. 1, 2

48. , 0.4

49. 6, 0.5

50. , 1.2

52. a. 4, 2b. y = 4 sin c. coil B

53. y = sin 60

54. y = sin 30

55. y = sin 240,000

56. 2 , 1

57. 2 , 1

58. , 1

59. a. days from spring equinox, hours of sunlight

b. h, about 365 days

c. y = sin

d. 1.1 h

e. Check students’ work.

2 x365

64. [2] Since sine is always positive in the first and second quadrants, a value of where its sine is equal to the sin 60° would be 180° – 60°, or 120°.

65. [4] The amplitude is 1, so a = 1.

30° • = radians,

so b = 2 ÷ = 2 • = 12.

The function is y = sin 12 .[3] one calculation error[2] incomplete explanation[1] answer only, with no work shown

66. – radians, –1.40 radians

67. radians, 2.62 radians

radians180° 6

69. radians, 5.59 radians

71. 49%

72. an = 7 – n; –8

73. an = 12 + 3n; 57

74. an = 0.8n – 0.2; 11.8

1. Sketch the graph of y = 3 sin 2 in the interval from 0 to 2 .

2. Write an equation of the sine function for this graph.

3. What is the amplitude of the graph in Question 2?13

y = sin13

The Cosine FunctionThe Cosine Function

(For help, go to Lesson 13-1 and 13-4.)

Find the x-coordinate of each point on the unit circle.

1. A 2. B

3. C 4. D

The Cosine FunctionThe Cosine FunctionALGEBRA 2 LESSON 13-5ALGEBRA 2 LESSON 13-5

1. x-coordinate of point A: 1

2. x-coordinate of point B: 0

3. x-coordinate of point C: –1

4. x-coordinate of point D: 0

Solutions

Use the graph shown below.

a. Find the domain, period range, and amplitude of this function.

The domain of the function is all real numbers.

The function goes from its maximum value of 2 and back again in an interval from 0 to 2 . The period is 2 .

The amplitude is (maximum – minimum) = (2 – (–2)) = (4) = 2.12

The function has a maximum value of 2 and a minimum value of –2. The range is –2 y 2.<– <–

(continued)

The maximum value occurs at 0 and 2 .

b. Examine the cycle of the cosine function in the interval from 0 to 2 . Where in the cycle does the maximum value occur? Where does the minimum occur? Where do the zeros occur?

The minimum value occurs at .

The zeros occur at and at .2 23

Divide the period into fourths. Plot five points for the first cycle. Use 2 for the maximum and –2 for the minimum.

Repeat the pattern for the second cycle.

Sketch the graph of y = –2 cos in the interval from 0 to 4.

| a | = 2, so the amplitude is 2.

b = , so the graph has 2 full cycles from 0 to 4. = 2, so the period is 2.2

Sketch the curve.

Suppose 8-in. waves occur every 6 s. Write an equation that

models the height of a water molecule as it moves from crest to crest.

= 4 Simplify.

a = amplitude = 82

maximum – minimum2

period = Use the formula for the period.2b

6 = The period is 6. Substitute.2b

b = Multiply each side by .b6

= Simplify.3

The equation will have the form y = a cos b . Find the values for a and b.

An equation that models the height of the water molecule is y = 4 cos . 3

Step 1: Use two equations. Graph the equations y = 1 and y = –2 cos on the same screen.2x

In the function y = –2 cos , for which values of x is the

function equal to 1?

Solve the equation, 1 = –2 cos , for the interval of 0 to 10.2x3

The graph show two solutions in the interval. They are x = 3.14 and 6.28.

Step 2: Use the Intersect feature to find the points at which the two graphs intersect.

The solution to the equation for the interval 0 x 10 is 3.14 and 6.28, or and 2 .

<– <–

1. 2 , 3; max: 0, 2 ; min: ; zeros: ,

2. , 1; max: 0, , , 2 ;

min: , , ; zeros: , , , , ,

3. , 1; max: 0, , 2 ; min: , ;

zeros: , , ,

4. 2 , 2; max: ; min: 0, 2 ; zeros: ,

35 3 6 2

10. y = 2 cos 2

11. y = cos

12. y = cos

13. y = –3 cos 2

14. y = 2 cos

15. y = 4 cos

16. 0.52, 2.62, 3.67, 5.76

17. 1.98, 4.30

18. 0.55, 1.45, 2.55, 3.45, 4.55, 5.45

19. 2.52

20. 0.00

21. 0.86, 5.14

22. 2 , –3 y 3, 3

23. , –1 y 1, 1

24. 4 , –2 y 2, 2

25. 4 , – y ,

26. 6 , –3 y 3, 3

27. , – y ,

28. , –16 y 16, 16

29. 2, –0.7 y 0.7, 0.7

30. 0.64, 2.50

31. 1.83, 2.88, 4.97, 6.02

32. 0.50, 2.50, 4.50

<– <–

33. a. 3.79, 5.64b. 10.07, 11.92; these values

are the sums of the values from part (a) and 2 .

34. a.

b. Answers may vary. Sample: 0 s, 4 s, 8 s, 12 s

c. 2 s; 2 s

35. a. 5.5 ft; 1.5 ftb. about 12 h 22 min

c. y = 1.5 cos

d. anytime except between 7:49 A.M. and 12:39 P.M.

36. a. 4 s; 6 ftb. y = –6 cos tc.

5d. –6 cos t = 3

e. No; at 13.5 s you are right of the puddle and moving to the right.

2 t742

37. a.

shift of units to the rightb.

They are the same.c. To write a sine function as a

cosine function, replace sin with cos and replace with – .

38. y = cos x or y = –cos x

39. On the unit circle, the x-values of – are equal to the x-values of , so cos(– ) = cos . –cos is the opposite of cos , so these graphs are reflections of each other over the x-axis.

y = sin 6

y = 2.5 sin 2

y = 4 sin 2

49. about 1111

50. about 204

51. about 83

52. an = 10 • 3n – 1; 10, 30, 90, 270, 810

53. an = 12(–0.3)n – 1; 12, –3.6, 1.08, –0.324, 0.0972

54. an = 900 – n – 1; 900, –300, 100, – ,13

1. Sketch the graph of y = 2 cos 2 in the interval from 0 to 6.

2. What is the amplitude of the graph in Question 1?

3. Write an equation of the cosine function for this graph.

y = 3 cos2x3

The Tangent FunctionThe Tangent Function

(For help, go to Lessons 13-4 and 13-5.)

Use a calculator to find the sine and cosine of each value of . Then calculate the ratio .

1. radians 2. 30 degrees

3. 90 degrees 4. radians

5. radians 6. 0 degrees

sin cos

The Tangent FunctionThe Tangent FunctionALGEBRA 2 LESSON 13-6ALGEBRA 2 LESSON 13-6

1. sin 0.866; cos = 0.5; 1.73

2. sin 30° = 0.5; cos 30° 0.866; 0.58

3. sin 90° = 1; cos 90° = 0; = , undefined

sin 3cos 3

0.8660.5

sin 30°cos 30°

0.50.866

sin 90°cos 90°

Solutions

4. sin = 0.5; cos –0.866; –0.58

5. sin = 1; cos = 0; = , undefined

6. sin 0° = 0; cos 0° = 1; = = 0

0.5–0.866

sin 0°cos 0°

Use the graph of y = tan to find each value.

a. tan –45° tan –45° = –1

b. tan 0° tan 0° = 0

c. tan 45° tan 45° = 1

Sketch the asymptotes.

Sketch two cycles of the graph y = tan .

period = Use the formula for the period.b

= = 2 Substitute for b and simplify.121

One cycle occurs in the interval – to .

Plot three points in each cycle.

Sketch the curve.

Asymptotes occur every 2 units, at = – , , and 3 .

What is the height of the triangle, in the design from

Example 3, when = 18°? What is the height when = 20°?

Step 1: Sketch the graph. Step 2: Use the TABLE feature.

When = 18°, the height of the triangle is about 32.5 ft.

When = 20°, the height of the triangle is about 36.4 ft.

3. –1

4. undefined

8. undefined

11. , = – , 5 10 10

12. , = – ,

13. , = – ,

14. , = – ,

2 3 3 3

8 843 2

50, undefined, –50

–100, undefined, 10021.

51.8, 125, 301.8 22. a.

b. 14.3 ftc. 20.2 ft

26. 1.11, 4.25

27. 2.03, 5.18

28. 0.08, 1.65, 3.22, 4.79

29. a.

b. Check students’ work; doubling the coefficient of the tangent function also doubles the output.

c. Answers may vary. Sample: the values of y = 600 tan x will be three times greater than the values of y = 200 tan x.

30. a. 140.4 ft2

1.7 in., 5.2 in.c. 5.2 in.2, 15.6 in.2

d. 3888 tiles, 1296 tiles

32. The asymptotes occur at x = –

and x = ; adding or subtracting

multiples of their difference, b, will

give other asymptote values.

33. 200

35. 135

36. –162

37. 70

38. y = tan x

39. y = –tan x

40. y = –tan x or

y = tan (–x)

41. y = tan (2x)

42. a.

6.9 ftb. 27.7 ft2

c. 166.3 ft2

43. a.

b. 130 ftc. 61,500 ft2

44. a. Check students’ work.b. The new pattern is asymptote

—(–a)—zero—(a)—asymptote.

45. Answers may vary. Sample: Triangles OAP and OBQ both share the angle and each triangle has a right angle, so

they are similar by AA. =

= = . Thus

= tan .

sin cossin cos

51. [2] There is no discussion of the amplitude of the tangent function because the tangent function has no max. or min. value.

52. [4] (Student graphs y = tan and y = sin correctly, showing 2 periods of y = tan and 1 period of y = sin .) The period of y = tan is half the period of y = sin .

[3] statement correct and graph accurate, but doesn’t illustrate 2 periods ofy = tan

[2] accurate graph with incorrect answer OR inaccurate graph with correct answer

[1] answer only, with no graph

53. 1.32, 4.97

54. 1.77, 4.51

55. 1.93, 4.35

56. 6.15

57. 2.95, 5.43

58. 0.44, 1.56, 2.44, 3.56, 4.44, 5.56

59. 5.9, 6, 4 and 6

60. 42.6, 42, 42

61. 8.0, 8, 7.9 and 8.5

62. 83

63. –227

64. 66

65. –8.3

66. 145

67. –332

1. Sketch the graph of y = tan 2 in the interval from 0 to 2 .

2. What is the period of the graph in Question 1?

3. If you know that the function y = tan x is undefined for x = a, what does that tell you about the graph at the value a?

The graph tends toward the vertical asymptote x = a.

Translating Sine and Cosine FunctionsTranslating Sine and Cosine Functions

Graph each pair of equations on the same coordinate plane. Identify each translation as horizontal, vertical, or diagonal.

1. y = 2x, y = 2x + 5 2. g(x) = | x |, ƒ(x) = | x + 3 |

3. y = –x, y = –x – 1 4. g(x) = | x |, h(x) = | x | – 4

5. y = –| x |, y = –| x – 2 | + 1 6. y = x2, y = (x + 3)2 – 2

Translating Sine and Cosine FunctionsTranslating Sine and Cosine FunctionsALGEBRA 2 LESSON 13-7ALGEBRA 2 LESSON 13-7

1. y = 2x 2. g(x) = | x |y = 2x + 5 f(x) = | x + 3|vertical or horizontalhorizontal translationtranslation

3. y = –x 4. g(x) = | x |y = –x – 1 h(x) = | x | – 4horizontal, verticalvertical, or translationdiagonaltranslation

Solutions

5. y = –| x | 6. y = x2

y = –| x – 2| + 1 y = (x + 3)2 – 2 diagonal diagonaltranslation translation

What is the value of h in each translation? Describe each

phase shift.

a. g(x) = ƒ(x + 3) b. y = | x – 2 |

h = –3; h = 2;

the phase shift is 3 units to the left.

the phase shift is 2 units to the right.

Use the graph of y = sin x. Sketch each translation of the

graph in the interval 0 x 2 .

a. y = sin x + 2

Translate the graph y = sin x 2 units up.

b. y = sin (x – )

Translate the graph y = sin x units to the right.

<– <–

Use the graph of y = sin x. Sketch the translation

y = sin – 1 in the interval 0 x 2 .x + 2

Translate the parent function units to

the left and 1 unit down.

<– <–

Step 2: Since h = and k = – , translate the graph units to the right and

unit down.

Extend the period pattern from 0 to 2 . Sketch the graph.

Sketch the graph of y = sin 3 – in the interval 0 to 2 .x – 2

Since a = 1 and b = 3, the graph is a translation of y = sin 3x.

Step 1: Sketch one cycle of y = sin 3x. Use five points in the pattern zero-max-zero-min-zero.

Write an equation for each translation.

Up means use a plus sign. Right means use a minus sign.

b. y = –sin x, 3 units to the righta. y = cos x, units up

A shift left or right means h = 3.A shift up or down means k = .

An equation is y = –sin (x – 3).An equation is y = cos x + .

Use the following graph, which shows the model for the data

given in Example 6, to draw some conclusions about the weather in

New Orleans.

One can draw the conclusion that the temperature for New Orleans is not steady, but fluctuates throughout the year.

More exactly that the temperature varies approximately through 48 degrees.

One can also determine that the hottest temperature for the city never gets above 100 degrees.

Nor does the coldest temperature ever get lower than 50 degrees.

1. –1; 1 unit to the left

2. –2; 2 units to the left

3. 1.6; 1.6 units to the right

4. 3; 3 units to the right

5. – ; units to the left

6. ; units to the right

17. 3, 2 ; 1 unit up

18. 4, ; 1 unit left and 2 units down

19. 1, 2 ; units left and 2 units up

20. 1, 2; 3 units right and 2 units up

31. y = sin (x + )

32. y = cos x –

33. y = sin x + 3

34. y = cos (x – 1.5)

35. y = cos

36. y = sin x – 3

42. y = –10 cos ; y = 10 sin x –

43. a. ; sin x = cos x –

b. – ; cos x = sin x +

44. a. 14.5 sin (x – 105.75) + 76.5

b. The difference between the two models is the horizontal shift.

c. about 66°Fd. March 20 (day 79)

45. a. Check students’ work.b. g(x) = ƒ(x + 4) – 3

46. a. y = 3 sin 2(x – 2) + 1b. 3, ; 2 units right and 1 unit up

37. a.

b. y = 8.5 cos (x – 228) + 77.5

38. y = sin (x – 2) – 4

39. y = cos (x + 3) +

40. y = sin x – + 3.5

41. y = 2 cos x – – 1;

y = 2 sin x + – 1

10 210

58. [2] If the function y = cos is

shifted to the right by

radians, the result is

y = cos – , which is

the same as y = sin .

So a = 1, and b = – .

59. [2] y = 4 sin x has an amplitude of 4, but a min. value of –1. Shift the graph up 5 units so the min. value is 4. The function is y = 4 sin + 5.

60. [4] sin = 3 sin

sin – sin = 3 sin – sin

0 = 2 sin

0 = sin

sin = 0 at – , 0, , and 2[3] one calculation error[2] incomplete answer[1] answer only, with no work shown

61. ; = – ,

62. 4 ; = –2 , 2

6 12 12

2 sin 2

63. ; = – ,

64. 6 ; = –3 , 3

65. 0.0064

66. 0.3456

67. 0.136

68. 0.198

69. 62

70. 16,383

71. –335,923

72. 96.09375

1. Sketch the graph of y = sin x – 1 in the interval from 0 to 2 .

2. What is the value of h in the translation g(x) = ƒ(x + 4)?

3. Describe the phase shift in the translation g(x) = ƒ(x + 4).

4. Write an equation for the translation of y = sin x, units down.

h = –4

four units to the left

y = sin x – 2

Reciprocal Trigonometric FunctionsReciprocal Trigonometric Functions

Graph each pair of functions on the same coordinate plane.

6. y = x, y = –x 7. y = x2, y = ± x

8. y = | 2x |, y = –| 2x | 9. y = –6x2, y = ± 6x

Find the reciprocal of each fraction.

1. 2. – 3. 4. 5. –58

1. The reciprocal of is . 2. The reciprocal of – is – .

3. The reciprocal of is or 2 . 4. The reciprocal of is .

5. The reciprocal of – is – .

6. y = x y = –x

7. y = x2 y = ± x

Solutions

8. y = | 2x| y = –| 2x|

9. y = –6x2 y = ± 6x

Reciprocal Trigonometric FunctionsReciprocal Trigonometric FunctionsALGEBRA 2 LESSON 13-8ALGEBRA 2 LESSON 13-8

a. Find csc 45°.

Use a calculator in degree mode.

Use the definition.

csc 45° = sin 45°1

b. Suppose cos = . Find sec .45

= Substitute.45

= Simplify.54

sec = Use the definition.cos 1

Find the exact value of csc 45°.

Use the unit circle to find the exact value of sin 45°. Then write the reciprocal.

The y-coordinate of point P is . 22

csc 45° = Use the definition.1

sin 45°

= Substitute.22

= Simplify.2

= Rationalize the denominator.2

Evaluate each expression. Use your calculator’s radian

mode. Round to the nearest thousandth.

a. cot = = 51

b. sec (–2) =1

cos (–2)

cot 1.3765

sec (–2) –2.403

Graph y = cos x and y = sec x in the interval from 0 to 2 .

Step 1: Make a table of values.

6 3 2 32

611 2x 0

cos x 1 0.9 .5 0 –.5 –0.9 –1 –0.9 –.5 0 .5 0.9 1

sec x 1 1.2 2 – –2 –1.2 –1 –1.2 –2 – 2 1.2 1

Step 2: Plot the points and sketch the graphs.

y = sec x will have a vertical asymptotewhenever its denominator (cos x) is 0.

Use the graph of y = sec x to find the value of sec 13°.

1cos xStep 1: Use degree mode. Graph y = .

Step 2: Use the TABLE feature.

sec 13° 1.0263

Use the function y = 6 sec to find the length of string needed

to form an angle of 55°.

Graph the function. Use the Value feature.

To form an angle of 55°, the string must be about 10.5 ft long.

y = 6 sec = 6 = Use the definition of secant.Simplify.

1. 1.02

2. 1.02

3. –0.70

4. –1.06

7. –

8. –

341813

12. –1

13. undefined

15. undefined

17. undefined

21. –7.02

22. –1

24. –1

25. –1.25

26. 17.13

27. 1.73

28. 1.02

33. 1.1547

34. 5.7588

35. –2.9238

37. 1.0642

38. 1.3054

39. 1.7321

40. 0.5774

41. a.

b. 28.3 ftc. 23.1 ftd. 20.7 ft

42. –1

43. ; 1.152 3 3

44. –1

45. –1

54. a. domain: all real numbers except multiples of ; range: all real numbers 1 or –1; period 2

b. 1c. –1

55. a. Reciprocals have the same sign.b. The reciprocal of –1 is –1.

56. csc 180° is undefined because

sin 180° is 0 and csc = .

57. sec 90° is undefined because

cos 90° is 0 and sec = .

58. cot 0° is undefined because

tan 0° is 0 and cot = .

>– <–

59. a.

b. 63.9 ftc. 69.3 ftd. 41.4°; 60.9 ft

60. a.

b. The domain of y = tan x is all real numbers except odd

multiples of , which are its

asymptotes. The domain of y = cot x is all real numbers except multiples of , which are its asymptotes. The range of both functions is all real numbers.

c. The graphs have the same period and range. Their asymptotes

are shifted by .

d. Answers may vary. Sample:

x = , x =

3 units up

units left

1 unit down

units right 2

4 units right

2 units left, 1 unit down

b. y = –cos x—domain: all real numbers; range: all real numbers between –1 and 1,inclusive; period: 2 ; y = –sec x—domain:

all real numbers except odd multiples of ; range: all real numbers except those between –1 and 1; period: 2

c. Multiples of ; sec (n ) = = =

±1 = cos (n ).d. Answers may vary. Sample: The graphs

have the same period and their signs are always the same. However, they have no range values in common except 1 and –1.

e. The signs of –sec x and –cos x are the same because reciprocals have the same sign.

units left, 3 units up68.

units right, 2 units down69. a.

1 cos (n )

70. a. IIb. I

71. y = sec x and y = csc x are not parabolas because parabolas are not restricted by asymptotes, whereas the branches of y = sec x and y = csc x are between asymptotes.

72. y = cos 3x has 3 cycles for each cycle of y = cos x. Thus, for each cycle of y = sec x, y = sec 3x has 3 cycles. Each

cycle of y = sec 3x is as wide

as one cycle of y = sec x.

73. a.

b. Answers may vary. Sample: Given y = cot bx, as |b| decreases, the period increases; as |b| increases, the period decreases. If b < 0, cot x begins each branch with negative y-values and ends with positive y-values; the opposite is true for b > 0.

74. a.

b. Answers may vary. Sample: Given y = b sec x, as |b| increases, the branches move further from the x-axis, but the asymptotes do not change. If b < 0, it is a reflection in the x-axis of y = |b| sec x.

79. 1.35

80. 0.6

81. 1.25

82. 3.4

83. 2, 2 ; 5 units down

84. 1, 2 ; 4 units left, 7 units down

85. 3, 2 ; units left, 4 units up

86. 5, 2; 1.5 units right, 8 units down

419 9 404140

89. a. 1(0) + 1(0) + 1(3); 3 units2

b. 1(3) + 1(3) + 1(12); 18 units2

90. a. 1(4) + 1(4) + 1(5); 13 units2

b. 1(5) + 1(5) + 1(8); 18 units2

91. a. 1(3) + 1(3) + 1(0); 6 units2

b. 1(4) + 1(4) + 1(3); 11 units2

92. a. 1(7) + 1(7) + 1(4); 18 units2

b. 1(8) + 1(8) + 1(7); 23 units2

1. What are the equations of the two lines that define the maximum and minimum values for the graphs of y = sin x and y = cos x?

2. Which basic trigonometric functions have vertical asymptotes?

3. What is the range of y-values that y sec x and y csc x cannot have?

y = 1 and y = –1

y = tan x, y = cot x,y = sec x, y = csc x

–1 < y < 1

Periodic Functions and TrigonometryPeriodic Functions and TrigonometryALGEBRA 2 CHAPTER 13ALGEBRA 2 CHAPTER 13

1. periodic; 4, 2

2. not periodic

3. 328°

4. 131°

5. 15°

6. – , –3.93

7. , 2.09

8. , 10.47

9. 150°

10. –450°

11. 46°

12. 1 cycle; 2, 2

13. cycle; 3, 4

15. Answers may vary. Sample: Multiply the radian measure

by . Example:

radians • = 120°

16. 42 in.

17. 4,

18. 2,

19. , ; 1.05, 5.24

20. , ; 1.05, 2.09

21. , , , ;

0.26, 1.83, 3.40, 4.97

29. y = sin x – 1

30. y = cos (x – 7.5)

31. y = sin (x + 3) – 1.5

32. y = cos x – + 8

37. undefined

121212

40. –

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